TSTP Solution File: CSR004+2 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : CSR004+2 : TPTP v8.1.2. Bugfixed v3.1.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof

% Computer : n001.cluster.edu
% Model    : x86_64 x86_64
% CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory   : 8042.1875MB
% OS       : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit  : 300s
% DateTime : Wed Aug 30 21:40:56 EDT 2023

% Result   : Theorem 0.20s 0.51s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13  % Problem  : CSR004+2 : TPTP v8.1.2. Bugfixed v3.1.0.
% 0.00/0.14  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35  % Computer : n001.cluster.edu
% 0.13/0.35  % Model    : x86_64 x86_64
% 0.13/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35  % Memory   : 8042.1875MB
% 0.13/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35  % CPULimit : 300
% 0.13/0.35  % WCLimit  : 300
% 0.13/0.35  % DateTime : Mon Aug 28 13:37:35 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.20/0.51  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.20/0.51  
% 0.20/0.51  % SZS status Theorem
% 0.20/0.51  
% 0.20/0.51  % SZS output start Proof
% 0.20/0.51  Take the following subset of the input axioms:
% 0.20/0.51    fof(filling_3, lemma, holdsAt(filling, n3)).
% 0.20/0.51    fof(happens_all_defn, axiom, ![Event, Time]: (happens(Event, Time) <=> ((Event=tapOn & Time=n0) | (holdsAt(waterLevel(n3), Time) & (holdsAt(filling, Time) & Event=overflow))))).
% 0.20/0.51    fof(overflow_3, conjecture, happens(overflow, n3)).
% 0.20/0.51    fof(waterLevel_3, lemma, holdsAt(waterLevel(n3), n3)).
% 0.20/0.51  
% 0.20/0.51  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.51  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.51  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.51    fresh(y, y, x1...xn) = u
% 0.20/0.51    C => fresh(s, t, x1...xn) = v
% 0.20/0.51  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.51  variables of u and v.
% 0.20/0.51  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.51  input problem has no model of domain size 1).
% 0.20/0.51  
% 0.20/0.51  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.51  
% 0.20/0.51  Axiom 1 (filling_3): holdsAt(filling, n3) = true2.
% 0.20/0.51  Axiom 2 (waterLevel_3): holdsAt(waterLevel(n3), n3) = true2.
% 0.20/0.51  Axiom 3 (happens_all_defn_1): fresh59(X, X, Y, Z) = true2.
% 0.20/0.51  Axiom 4 (happens_all_defn_1): fresh58(X, X, Y, Z) = fresh59(Y, overflow, Y, Z).
% 0.20/0.51  Axiom 5 (happens_all_defn_1): fresh53(X, X, Y, Z) = happens(Y, Z).
% 0.20/0.51  Axiom 6 (happens_all_defn_1): fresh58(holdsAt(waterLevel(n3), X), true2, Y, X) = fresh53(holdsAt(filling, X), true2, Y, X).
% 0.20/0.51  
% 0.20/0.51  Goal 1 (overflow_3): happens(overflow, n3) = true2.
% 0.20/0.51  Proof:
% 0.20/0.51    happens(overflow, n3)
% 0.20/0.51  = { by axiom 5 (happens_all_defn_1) R->L }
% 0.20/0.51    fresh53(true2, true2, overflow, n3)
% 0.20/0.51  = { by axiom 1 (filling_3) R->L }
% 0.20/0.51    fresh53(holdsAt(filling, n3), true2, overflow, n3)
% 0.20/0.51  = { by axiom 6 (happens_all_defn_1) R->L }
% 0.20/0.51    fresh58(holdsAt(waterLevel(n3), n3), true2, overflow, n3)
% 0.20/0.51  = { by axiom 2 (waterLevel_3) }
% 0.20/0.51    fresh58(true2, true2, overflow, n3)
% 0.20/0.51  = { by axiom 4 (happens_all_defn_1) }
% 0.20/0.51    fresh59(overflow, overflow, overflow, n3)
% 0.20/0.51  = { by axiom 3 (happens_all_defn_1) }
% 0.20/0.51    true2
% 0.20/0.51  % SZS output end Proof
% 0.20/0.51  
% 0.20/0.51  RESULT: Theorem (the conjecture is true).
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